Find at least 4 consecutive values less than 1 in an array
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I have an array in which I need to find four or more consecutive values that are less than one. For example: M = [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3]
The program should return the indexes 3-6 and 13-18.
How can I go about solving this?
1 Kommentar
José-Luis
am 29 Jan. 2014
That would be 2-6 and 13-18
Antworten (6)
Roger Stafford
am 29 Jan. 2014
Bearbeitet: Roger Stafford
am 29 Jan. 2014
Here's another one-liner if you accept initial indices as an answer.
f = find(all(hankel(M(1:4),M(4:end))<1,2));
1 Kommentar
Jos (10584)
am 29 Jan. 2014
+1
Azzi Abdelmalek
am 29 Jan. 2014
Bearbeitet: Azzi Abdelmalek
am 29 Jan. 2014
Edit
M= [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3]
if diff(size(M))<0
M=M'
end
a=M<1;
b=strrep(num2str(a),' ','');
[ii,jj]=regexp(b,'1+','start','end')
d=jj-ii>=4
out=[ii;jj];
out=out(:,d)
3 Kommentare
Neha
am 29 Jan. 2014
Azzi Abdelmalek
am 29 Jan. 2014
Bearbeitet: Azzi Abdelmalek
am 29 Jan. 2014
If you copy and paste the code, there is no error, unless you are using M that is not a 1xn array. What is the size of M
size(M)
Azzi Abdelmalek
am 29 Jan. 2014
If M is a column array, just add at the beginning of your code
M=M'
Azzi Abdelmalek
am 29 Jan. 2014
Bearbeitet: Azzi Abdelmalek
am 29 Jan. 2014
Edit
M= [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3];
if diff(size(M))<0
M=M'
end
a=[0 M<1 0];
ii=strfind(a,[0 1]);
jj=strfind(a,[1 0])-1;
d=jj-ii>=4;
out=[ii;jj];
out=out(:,d)
3 Kommentare
Neha
am 29 Jan. 2014
Azzi Abdelmalek
am 29 Jan. 2014
It doesn't matter if the length changes. Maybe you are asking if the size changes, from 1xn to nx1. In this case we can edit our code. Look at edited answesrs
Neha
am 29 Jan. 2014
José-Luis
am 29 Jan. 2014
M = [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3];
numVal = 4;
your_vals = conv(double(M<1),ones(1,numVal)/numVal,'valid') == 1;
start = find(diff([0 your_vals]) == 1)
finish = fliplr(numel(M) + 1 - find(diff([0 fliplr(your_vals)]) == 1))
Andrei Bobrov
am 29 Jan. 2014
t = M<1&M>0;
t1 = [true;diff(t(:))~=0];
idx=accumarray(cumsum(t1),(1:numel(M))',[],@(x){x});
ii = idx(t(t1));
out = ii(cellfun(@numel,ii)>=4);
Jos (10584)
am 29 Jan. 2014
Bearbeitet: Jos (10584)
am 29 Jan. 2014
Here is a relatively simple one-liner:
M = [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3] % example data
[startIDX, endIDX] = regexp(char((M(:).'< 1)+'0'), '1111+')
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